Velocity vs displcement graph

[hav0c]

i was wondering what the equation of the line/curve on a velocity vs displacement graph would be which would indicate constant acceleration.
I am totally stumped.
EDIT:please feel free to shift the thread to any other place if need be, but since it wasn't coursework this seemed to be the natural place.

 

[dipole]

Well you know that in general the equation of motion for a particle of constant acceleration is given by:

x(t) = \frac{1}{2}at^2 + v_0t + x_0

and

v(t) = at + v_0


So, solving the first equation for t:

\frac{1}{2}at^2 + v_0t + (x_0 -x) = 0
t = \frac{-v_0 \pm \sqrt{v_0^2 - 2a(x_0-x)}}{a}

Now plug that into the equation for v:

v(x) = \pm \sqrt{v_0^2 - 2a(x_0-x)}

I skipped some steps, so I encourage you to work the math out yourself, it's quite simple. Do you understand why the velocity changes slower and slower as the displacement increases? Once you see that, then you'll understand why it had to have that shape even without doing the math.

 

[hav0c]

v(x) = \pm \sqrt{v_0^2 - 2a(x_0-x)}

isn't this just v^2-u^2=2ax?

Anyways, i see that with increase in x, a decreases
therefore the graph looks like this?
(see attachment)

 

[dipole]

Yes it look like that, but the velocity doesn't decrease with x, the rate of change of the velocity decreases with x. Not that this is not acceleration, because that is the change of velocity with respect to time.

 

[dodi]

then what is the name for this curve ? and is this the graph for velocity versus displacement ?

 

[Kajal Sengupta]

There is no separate name for the graph. It is simply the velocity vs displacement graph.

 

[sophiecentaur]

then what is the name for this curve ? and is this the graph for velocity versus displacement ?

Isn't it just a parabola? From school lessons y² = 4ax
It is possible to plot more or less any variable against any other variable in a process and it may or may not give a useful picture. 'Names' aren't really very important and only used for the more common ones.